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san4es73 [151]
2 years ago
3

On flat ground, a 70-kg person requires about 300 W of metabolic power to walk at a steady pace of 5.0 km/h (1.4 m/s). Using the

same metabolic power output, that person can bicycle over the same ground at 15 km/h.A 70-kg person walks at a steady pace of 5.0 km/h on a treadmill at a 5.0% grade. (That is, the vertical distance covered is 5.0% of the horizontal distance covered.) If we assume the metabolic power required is equal to that required for walking on a flat surface plus the rate of doing work for the vertical climb, how much power is required?
Physics
1 answer:
VLD [36.1K]2 years ago
4 0

Answer:

The required power is 471.5W

Explanation:

Let d be the horizontal distance traveled on trade mil

Vertical distance traveled=0.05d

Time takes to travel distance d=d/5 hr  

The energy for vertical motion is given as:

E=mgh\\E=70kg(9.81m/s^{2} )0.05d\\E=34.3d

And the Power is given as:

Power=Energy/time\\Power=\frac{34.3d}{\frac{d}{5} }\\ Power=171.5W

So the total power is:

Power_{total} =300W+171.5W\\Power_{total} =471.5W

The required power is 471.5W

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2 years ago
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__________ curves help lessen the effect of the force of the forward motion on your vehicle as it enters the curve.
Rainbow [258]

Answer:

Banked

Explanation:

Banked curves are formed when the inner edge is below the outer edge.

It is done in order to ensure the reliability of the frictional force as it varies when the road is wet wet or oily. Thus in order to avoid these problems the curved roads are banked.

Banking of the curve provides the necessary centripetal force, i.e., the horizontal component of the normal reaction force to keep the vehicle i motion and thus helps in reducing the effect of the forward motion force on the vehicle.

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2 years ago
You have been hired to check the technical correctness of an upcoming made-for-TV murder mystery that takes place in a space shu
AlladinOne [14]

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The astronaut who has a mass of 80 kg without the toolkit do survive with 40 seconds of remaining air

Explanation:

Due the astronaut throws the 10-kg tool kit away with a speed of 8 m/s, it gives a momentum equivalent but in the other direction, so I=mv=(10Kg)(8m/s)=80kg*m/s, then we can find the speed that the astronaut reaches due to its weight we get, v=\frac{I}{m} =\frac{80kg*m/s}{80Kg} =1m/s.

Finally, as the distance to the space shuttle is 200m, the time taken to the astronaut to reach it at the given speed will be t=\frac{d}{v}=\frac{200m}{1m/s}=200s, as the remaining air time is 4 min or 240 seconds, The astronaut who has a mass of 80 kg without the toolkit do survive with 40 seconds of remaining air.

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2 years ago
A slingshot can project a pebble at a speed as high as 38.0 m/s. (a) If air resistance can be ignored, how high (in m) would a p
kipiarov [429]

Answer:

73.67 m

Explanation:

If projected straight up, we can work in 1 dimension, and we can use the following kinematic equations:

y(t) = y_0 + V_0 * t + \frac{1}{2} a t^2

V(t) = V_0 + a * t,

Where y_0 its our initial height, V_0  our initial speed, a the acceleration and t the time that has passed.

For our problem, the initial height its 0 meters, our initial speed its 38.0 m/s, the acceleration its the gravitational one ( g = 9.8 m/s^2), and the time its uknown.

We can plug this values in our equations, to obtain:

y(t) =  38 \frac{m}{s} * t - \frac{1}{2} g t^2

V(t) = 38 \frac{m}{s} - g * t

note that the acceleration point downwards, hence the minus sign.

Now, in the highest point, velocity must be zero, so, we can grab our second equation, and write:

0 m = 38 \frac{m}{s} - g * t

and obtain:

t = 38 \frac{m}{s} / g

t = 38 \frac{m}{s} / 9.8 \frac{m}{s^2}

t = 3.9 s

Plugin this time on our first equation we find:

y = 38 \frac{m}{s} * 3.9 s - \frac{1}{2} 9.8 \frac{m}{s^2} (3.9 s)^2

y=73.67 m

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taurus [48]
-3 m/s
---------
per min

oh I think 8m/s to 3m/s to 0m/s

idk probably -0.08 

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